Riemann integrable proof of a formula with partitions

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If $f:[a,b] \to \mathbb{R}$ is Riemann integrable and $F: [a,b] \to \mathbb{R}$ verifies $F' = f$, then

$$ F(b)-F(a) = \int_a^b f(t)dt $$

Hint: if $P=(x_0,x_1,...,x_n)$ is a partition of $[a,b]$, write

$$ F(b)-F(a) = \sum_{1 \leq i \leq n} F(x_i) - F(x_{i-1}) $$

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By the mean value theorem there are points $\xi_j \in (x_{j-1},x_j)$ such that

$$F(b) - F(a) = \sum_{j=1}^n [F(x_j) - F(x_{j-1})] = \sum_{j=1}^n F'(\xi_j)(x_j - x_{j-1}) = \sum_{j=1}^n f(\xi_j)(x_j - x_{j-1}). $$

Do you recognize the sum on the RHS?